a) 1/2 +1/4+1/8+…+1/128 Giúp mình nhé
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 5 2023
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)
A\(\times\) 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)
A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)
A = \(\dfrac{127}{128}\)
BD
28 tháng 5 2023
Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(B=1+0-\dfrac{1}{128}\)
\(B=1-\dfrac{1}{128}\)
\(B=\dfrac{128}{128}-\dfrac{1}{128}\)
\(B=\dfrac{127}{128}\)
NT
1
17 tháng 9 2021
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)
A
21 tháng 6 2018
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)
\(=\frac{1}{2}-\frac{1}{512}\)
\(=\frac{255}{512}\)
Vậy \(A=\frac{255}{512}\)
21 tháng 6 2018
=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152
=1/2-1/152
=255/512
A=255/512
NM
9
22 tháng 5 2016
1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64 + 1/128
= 64/ 128 + 32/128 + 16/128 +8/128 + 4/128 +2/128 + 1/128
= ( 64 + 32 + 16 + 8 + 4 + 2 + 1 ) /128
= 127/ 128
22 tháng 5 2016
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - ............ + 1/64 - 1/128
= 1 - 1/128
= 127/128
k nha bn
T
9 tháng 1
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dots+\dfrac{1}{32}+\dfrac{1}{64}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(A=1-\dfrac{1}{128}=\dfrac{127}{128}\)
LG
2
7 tháng 4 2022
1/2 + 1/4 + 1/8 + … + 1/128
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + … + 1/64 - 1/128
= 1 - 1/128
= 128/128 - 1/128
= 127/128
Chúc bạn học tốt.
😁😁😁
TH
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 5 2023
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)
A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)
A \(\times\)( 2-1) = \(\dfrac{255}{128}\)
A = \(\dfrac{255}{128}\)
BD
29 tháng 5 2023
Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T
\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)
\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(T=2+0+0+...-\dfrac{1}{128}\)
\(T=\dfrac{256}{128}-\dfrac{1}{128}\)
\(T=\dfrac{255}{128}\)
17 tháng 9 2021
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
1/2=64/128
1/4=32/128
...
Ta có:
\(\dfrac{1+2+4+...+64}{128}\)=\(\dfrac{127}{128}\)